We have three 12/220kV, 40MVA generator transformers with impedance voltage at 10.8% that are paralled. We have run out of capacity and the only other transformer immediately available is rated 18/330kV, 125MVA impedance voltage at 12%. As an interim measure can we connect this unit in parallel to the other three, seeing that the voltage ratio is the same? I mean can we connect this large unit on the 12kV bus (in parallel to the other three) to get 220kV without any problems? The winding connection is delta-star in both cases.
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paralleling power transformers with different impedances 1
- Thread starter vidima
- Start date
Hi Vidima
In your case you have remember two from four rules of trafo's paralleling.
1. Uk difference isn't more than 10% ( probably you are on the limit 10.8+10%=11.88%).
2. Difference of Power of trafo's isn't more than 3:1.
125/40> 3 ( but not so much and this only in case of only one 40MVA trafo is connect to the bus).
That means, very critical, but possible
In additional you have calculate load share in three cases:
1. 1 trafo 40MVA and 125MVA
2. 2 trafo 40MVA and 125MVA
3. 3 trafo 40MVA and 125MVA
Best Regards.
Slava
In your case you have remember two from four rules of trafo's paralleling.
1. Uk difference isn't more than 10% ( probably you are on the limit 10.8+10%=11.88%).
2. Difference of Power of trafo's isn't more than 3:1.
125/40> 3 ( but not so much and this only in case of only one 40MVA trafo is connect to the bus).
That means, very critical, but possible
In additional you have calculate load share in three cases:
1. 1 trafo 40MVA and 125MVA
2. 2 trafo 40MVA and 125MVA
3. 3 trafo 40MVA and 125MVA
Best Regards.
Slava
You should also look at the X:R ratios, and the %R.
The impedance determines the current under short circuit conditions. The resistance may predominate under load conditions.
With differing impedances (and differing resistances), the sum of the transformer currents can be expected to exceed the load current. (Differing current phase angles).
With differing impedances the transformers will not share the load proportionately. The transformer with the higher impedance will take a lesser share of the load.
You are probably safe, but be aware that when the three existing transformers are 100% loaded, the new transformer will be less than 100% loaded. Taking a holistic view, you have 40 MVA x 3 = 120 MVA and need more capacity. The new transformer may be re-rated to 10.8%Z.
125 MVA x (10.8%/12%) = 112.5 MVA. Your application is on the conservative side of the equation.
The transformer being added is large compared to the existing transformers and the per unit impedance is slightly greater.
Had the situation been reversed and you were to consider adding a smaller transformer of lesser per unit impedance, you could actually reduce the safe capacity of the combination rather than increase the capacity. The smaller transformer could be 100% loaded while the larger transformers were at less than 100% load.
You should be aware that the statement that transformers of equal per unit impedances will share the load in proportion to their respective capacities is based on the assumption that the X:R ratios are equal. This is often the case with the sizes of transformers that are normally called upon to work in parallel.
You are effectively adding 112.5 MVA to an existing 120 MVA of capacity. You are almost doubling your capacity. You should be safe.
If you were going to apply a load in the order of 230MVA to the combination it would be prudent to calculate the current sharing based on the X:R ratios and the actual X and R of the transformers combined with the power factor of the load (which is based on the X and R of the load).
Bill
--------------------
"Why not the best?"
Jimmy Carter
The impedance determines the current under short circuit conditions. The resistance may predominate under load conditions.
With differing impedances (and differing resistances), the sum of the transformer currents can be expected to exceed the load current. (Differing current phase angles).
With differing impedances the transformers will not share the load proportionately. The transformer with the higher impedance will take a lesser share of the load.
You are probably safe, but be aware that when the three existing transformers are 100% loaded, the new transformer will be less than 100% loaded. Taking a holistic view, you have 40 MVA x 3 = 120 MVA and need more capacity. The new transformer may be re-rated to 10.8%Z.
125 MVA x (10.8%/12%) = 112.5 MVA. Your application is on the conservative side of the equation.
The transformer being added is large compared to the existing transformers and the per unit impedance is slightly greater.
Had the situation been reversed and you were to consider adding a smaller transformer of lesser per unit impedance, you could actually reduce the safe capacity of the combination rather than increase the capacity. The smaller transformer could be 100% loaded while the larger transformers were at less than 100% load.
You should be aware that the statement that transformers of equal per unit impedances will share the load in proportion to their respective capacities is based on the assumption that the X:R ratios are equal. This is often the case with the sizes of transformers that are normally called upon to work in parallel.
You are effectively adding 112.5 MVA to an existing 120 MVA of capacity. You are almost doubling your capacity. You should be safe.
If you were going to apply a load in the order of 230MVA to the combination it would be prudent to calculate the current sharing based on the X:R ratios and the actual X and R of the transformers combined with the power factor of the load (which is based on the X and R of the load).
Bill
--------------------
"Why not the best?"
Jimmy Carter
-
1
- #5
slavag,
I think you're mixing a rule of thumb (10% max difference on impedance) that is really intended to be used with single phase transformers. The 10% difference doesn't matter here.
In this case, the transformer impedance on the lower voltage base is:
1) 0.389 ohms on 12 kV for the 40 MVA transformers
2) 0.311 ohms on 18 kV for the 125 MVA transformers
At this point you can treat the load sharing issues as parallel resistors for the 3 cases that you listed.
Judging by the impedance values, the 125 MVA transformer will take more of the load than the 40 MVA transformer(s). This is a good thing.
So, vidima, I don't see a problem with doing what you suggested because the voltage ratio is the same and the impedance values look like they will work. You can parallel the 3 - 40 MVA transformers with the 125 MVA transformer and not worry about anything. The 125 MVA transformer will carry more load than the 40 MVA transformers and it has more capability to do so.
I hadn't run through the numbers yet, but you can do as I suggested earlier and treat the transformers as parallel resistors.
I think you're mixing a rule of thumb (10% max difference on impedance) that is really intended to be used with single phase transformers. The 10% difference doesn't matter here.
In this case, the transformer impedance on the lower voltage base is:
1) 0.389 ohms on 12 kV for the 40 MVA transformers
2) 0.311 ohms on 18 kV for the 125 MVA transformers
At this point you can treat the load sharing issues as parallel resistors for the 3 cases that you listed.
Judging by the impedance values, the 125 MVA transformer will take more of the load than the 40 MVA transformer(s). This is a good thing.
So, vidima, I don't see a problem with doing what you suggested because the voltage ratio is the same and the impedance values look like they will work. You can parallel the 3 - 40 MVA transformers with the 125 MVA transformer and not worry about anything. The 125 MVA transformer will carry more load than the 40 MVA transformers and it has more capability to do so.
I hadn't run through the numbers yet, but you can do as I suggested earlier and treat the transformers as parallel resistors.
Hi Magoo2
Maybe I'm mix.
I use simple rule of tumb
S/Uk=Sr1/Uk1+Sr2/Uk2+..Srn/Ukn
from this we can see resultnt impedance
Uk=S/ ( Sr1/Uk1+Sr2/Uk2+...)
and after this power according to each trafo
S1= Sr1^Uk/Uk1
lets try calculate.
case 1. 3 trafos 40MVA and one 125MVA
Uk= 245MVA/( 120/10.8+125/12)=11.38
S1= 40x11.38/10.8= 42MVA ( oveload)
S2= 125x11.38/12= 118.5MVA ( underload)
case 2 two trafos and one 125MVA
205/(80/10.8+125/12)= 11.5
case 3 one 40 and one 125
165/(40/10.8+125/12)= 11.7
40x11.7/10.8= 43MVA overload.
This is result
Regards.
Slava
Maybe I'm mix.
I use simple rule of tumb
S/Uk=Sr1/Uk1+Sr2/Uk2+..Srn/Ukn
from this we can see resultnt impedance
Uk=S/ ( Sr1/Uk1+Sr2/Uk2+...)
and after this power according to each trafo
S1= Sr1^Uk/Uk1
lets try calculate.
case 1. 3 trafos 40MVA and one 125MVA
Uk= 245MVA/( 120/10.8+125/12)=11.38
S1= 40x11.38/10.8= 42MVA ( oveload)
S2= 125x11.38/12= 118.5MVA ( underload)
case 2 two trafos and one 125MVA
205/(80/10.8+125/12)= 11.5
case 3 one 40 and one 125
165/(40/10.8+125/12)= 11.7
40x11.7/10.8= 43MVA overload.
This is result
Regards.
Slava
Each 40 MVA transformer has an impedance of 0.389 ohms. The 125 MVA transformer has an impedance of 0.311 ohms.
Since these are generator transformers, assume the total MVA is 160 since you started with 40 MVA each.
If you convert to admittances, you'll find that each of the 3 - 40 MVA transformers will be carrying around 37.6 MVA and the 125 MVA transformer will be carrying 47.1 MVA. None of these will be overloaded.
It should be obvious from the impedances. The larger transformer will take more of the load than the 40 MVA transformers because 0.311 < 0.389.
Is my math wrong?
He's trying to replace a failed 40 MVA with 125 MVA. The generator ratings didn't change.
Since these are generator transformers, assume the total MVA is 160 since you started with 40 MVA each.
If you convert to admittances, you'll find that each of the 3 - 40 MVA transformers will be carrying around 37.6 MVA and the 125 MVA transformer will be carrying 47.1 MVA. None of these will be overloaded.
It should be obvious from the impedances. The larger transformer will take more of the load than the 40 MVA transformers because 0.311 < 0.389.
Is my math wrong?
He's trying to replace a failed 40 MVA with 125 MVA. The generator ratings didn't change.
Sorry, I missed the voltage difference. The new transformer will be run at less than rated voltage.
if the transformer impedance is rated at 12% at 330kV, it will be rated at 18%Z at 220 volts. It will take the same applied voltage to force full rated current through shorted windings.
That voltage is now a percentage of 220kV instead of 330 kV and the transformer will re-rate to 83.3 MVA.
If we want to consider the transformer as a 10.8% impedance rated transformer we must de-rate it further in the ratio of the impedances (10.8/18). Considered as a 10.8% impedance transformer the new rating is 50 MVA. Although the transformer is capable of delivering more than 50 MVA at 220kV, it will act as a 50 MVA transformer as far as load sharing.
I believe that these figures agree with Magoo2's calculations.
Different method, same result.
Bill
--------------------
"Why not the best?"
Jimmy Carter
if the transformer impedance is rated at 12% at 330kV, it will be rated at 18%Z at 220 volts. It will take the same applied voltage to force full rated current through shorted windings.
That voltage is now a percentage of 220kV instead of 330 kV and the transformer will re-rate to 83.3 MVA.
If we want to consider the transformer as a 10.8% impedance rated transformer we must de-rate it further in the ratio of the impedances (10.8/18). Considered as a 10.8% impedance transformer the new rating is 50 MVA. Although the transformer is capable of delivering more than 50 MVA at 220kV, it will act as a 50 MVA transformer as far as load sharing.
I believe that these figures agree with Magoo2's calculations.
Different method, same result.
Bill
--------------------
"Why not the best?"
Jimmy Carter
Hi Magoo2 and Bill.
I think we have some misunderstanding. Isn't good.
Maybe I miss something.
Please explain me two other cases.
1. One trafo is 50MVA with impedance 10% and second 150MVA with impedance 15%. Same voltage ratio and vector group
Are possible connect them in parallel and what will be load sharing?
2. Treee 50MVA trafos same voltage ratio and vector group, with impedance 8%,10% and 12%.
Are possible connect them in parallel and what will be load sharing?
Please believe me, I don't check you , I check myself only.
Isn't good that in this case we have different result.
Best Regards.
Slava
I think we have some misunderstanding. Isn't good.
Maybe I miss something.
Please explain me two other cases.
1. One trafo is 50MVA with impedance 10% and second 150MVA with impedance 15%. Same voltage ratio and vector group
Are possible connect them in parallel and what will be load sharing?
2. Treee 50MVA trafos same voltage ratio and vector group, with impedance 8%,10% and 12%.
Are possible connect them in parallel and what will be load sharing?
Please believe me, I don't check you , I check myself only.
Isn't good that in this case we have different result.
Best Regards.
Slava
Your math is the same as mine, except that you assumed 160 MVA total - probably a good assumption.
Although it is obvious from the impedances that the 125 MVA transformer will carry more load than a single 40 MVA transformer, it has a much higher rating. It is not obvious that the load is OK for the smaller transformers.
The OP is not replacing a failed 40 MVA with 125 MVA, he is adding a 125 MVA in parallel to 3 existing 40 MVA.
Magoo,
You method of converting to ohmic impedance and treating them as parallel resistors is a lot easier for me to understand than Slava's method of dividing power by voltage or Bill's method of finding equivalent power ratings. Your method could apply to different X/R ratios if you used complex impedances in the calculations.
You method of converting to ohmic impedance and treating them as parallel resistors is a lot easier for me to understand than Slava's method of dividing power by voltage or Bill's method of finding equivalent power ratings. Your method could apply to different X/R ratios if you used complex impedances in the calculations.
Bill., BTW.
I re-read your posts.
We have exactly same result.
lesser impedance - more load.
I calculated 125MVA with 12%, you 83.3 with 18%.
Your and "my" method is more or less same.
Load Share between trafos according to Uk.
that means if you have 1000kVA trafo and one with 3% and second with 6% impedance
3% take 1200kVA and 6% 600kVA, total load will be only 1800kVA.
Best Regards.
Slava
I re-read your posts.
We have exactly same result.
lesser impedance - more load.
I calculated 125MVA with 12%, you 83.3 with 18%.
Your and "my" method is more or less same.
Load Share between trafos according to Uk.
that means if you have 1000kVA trafo and one with 3% and second with 6% impedance
3% take 1200kVA and 6% 600kVA, total load will be only 1800kVA.
Best Regards.
Slava
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